CE 374 K – Hydrology
Runoff Processes
Daene C. McKinney
Watershed

Watershed



Area draining to a stream
Streamflow generated by
water entering surface
channels
Affected by




Physical, vegetative, and
climatic features
Geologic considerations
Stream Patterns
Dry periods

Flow sustained from
groundwater (baseflow)
http://www.epa.gov/owow/watershed/whatis.html
Streamflow
Atmospheric Moisture

Atmospheric Water





Rain
Evapotranspiration
Precipitation
Subsurface Water

Snow
Infiltration
Groundwater
Surface Water
Evaporation
Interception
Energy
Throughfall and
Stem Flow
Snowpack
Snowmelt
Pervious
Surface
Infiltration
Soil Moisture
Percolation
Groundwater
Evapotranspiration
Overland
Flow
Groundwater Flow
Streams and Lakes
Channel Flow
Runoff
Watershed
Boundary
Impervious
Evaporation
Streamflow Hydrograph
Basin Lag
Centroid of
Precipitation
Peak
Discharge, Q
Time
of Rise
Inflection
Point
Baseflow
Recession
Baseflow
Recession
Beginning of
Direct Runoff
End of
Direct Runoff
Time
Baseflow Separation

No inflow added to
groundwater - depletion
(recession) curve
Continuity equation
dS
 I (t )  Q(t )
dt
dS  Q0e (t t0 ) / k dt
Q(t )  Q0e (t t0 ) / k
Discharge, Q

Baseflow
Recession
S (t )  kQ(t )
Q(t )  flow at time t
Q0  flow at time t 0
k  decay constant [T ]
Time
Baseflow Separation Techniques
Straight – line method

Draw a horizontal line
segment (A-B) from
beginning of runoff to
intersection with
recession curve
Discharge, Q

Direct Runoff
B
A
Baseflow
Time
Baseflow Separation Techniques
Fixed Base Method


Draw line segment (A –
C) from baseflow
recession to a point
directly below the
hydrograph peak
Draw line segment (C-D)
connecting a point N time
periods after the peak
N  A0.2
Discharge, Q

Direct Runoff
A
C
Baseflow
D
B
Time
Baseflow Separation Techniques
Variable Slope Method



Draw line segment (A-C)
forward from baseflow
recession to a point
directly below the
hydrograph peak
Draw line segment (B-E)
backward from baseflow
recession to a point
directly below the
inflection point
Draw line segment (C-E)
Discharge, Q

A
Direct Runoff
E
C
Baseflow
B
Time
Abstraction (Losses) Estimation
Phi – Index Method

Excess (effective) rainfall




Abstraction (losses)


Rainfall that is not retained or
infiltrated
Becomes direct runoff
Excess rainfall hyetograph
(excess rainfall vs time)
Difference between total and
excess rainfall hyetographs
Phi – Index

Constant rate of abstraction
yielding excess rainfall
hyetograph with depth equal to
depth of direct runoff
M
rd   Rm  t 
m 1
rd  depth of direct runoff
Rm  observed rainfall
  Phi index
M  # intervals of rainfall
contributing to driect runoff
t  time interval
Example
Have precipitation and streamflow data, need to estimate losses
Observed
Rain
Flow
in
cfs
8:30
12000
203
9:00
0.15
246
9:30
0.26
283
10:00
1.33
828
10:30
2.2
2323
11:00
0.2
5697
11:30
0.09
9531
12:00
11025
12:30
8234
1:00
4321
0
0.5
10000
1
1.5
2
Streamflow (cfs)
Time
8000
2.5
6000
4000
2000
1:30
2246
2:00
1802
2:30
1230
3:00
713
3:30
394
4:00
354
4:30
303
0
7:30 PM
9:00 PM
10:30 PM
12:00 AM
1:30 AM
3:00 AM
Time
No direct runoff until after 9:30
And little precip after 11:00
Basin area A = 7.03 mi2
4:30 AM
6:00 AM
Example (Cont.)
Estimate baseflow (straight line method)

Constant = 400 cfs
12000
10000
Streamflow (cfs)

8000
6000
4000
2000
0
7:30 PM
9:00 PM
10:30 PM
12:00 AM
Time
1:30 AM
3:00 AM
4:30 AM
baseflow
6:00 AM
Example (Cont.)

Calculate Direct Runoff
Hydrograph

Subtract 400 cfs
Time
8:30
9:00
9:30
10:00
10:30
11:00
11:30
12:00
12:30
1:00
1:30
2:00
2:30
3:00
3:30
4:00
4:30
Observed
Rain
Flow
in
cfs
0.15
203
0.26
246
1.33
283
2.2
828
2.08
2323
0.2
5697
0.09
9531
11025
8234
4321
2246
1802
1230
713
394
354
303
Direct
Runoff
cfs
428
1923
5297
9131
10625
7834
3921
1846
1402
830
313
43550
Total = 43,550 cfs
Example (Cont.)

Compute volume of direct runoff
11
11
n 1
n 1
Vd   Qn t  t  Qn
 3600s/hr * 0.5 hr * 43,550 ft 3 /s
 7.839*107 ft 3

Compute depth of direct runoff
V
rd  d
A

7.839*107 ft 3
7.03 mi * 52802 ft 2
 0.4 ft
 4.80 in
Example (Cont.)


Neglect all precipitation intervals that occur before
the onset of direct runoff (before 9:30)
Select Rm as the precipitation values in the 1.5 hour
period from 10:00 – 11:30
M
rd   Rm  t 
m 1
4.80  (1.33  2.20  2.08   * 3 * 0.5)
  0.54 in
t  0.27 in
rd  4.80 in
Example (Cont.)
8:30
9:00
9:30
10:00
10:30
11:00
11:30
12:00
12:30
1:00
1:30
2:00
2:30
3:00
3:30
4:00
4:30
Observed
Rain
Flow
in
cfs
0.15
203
0.26
246
1.33
283
2.2
828
2.08
2323
0.2
5697
0.09
9531
11025
8234
4321
2246
1802
1230
713
394
354
303
Direct
Runoff
Excess
Rainfall
cfs
in
12000
0
t=0.27
0.5
428
1923
5297
9131
10625
7834
3921
1846
1402
830
313
43550
10000
1.06
1.93
1.81
1
1.5
2
Streamflow (cfs)
Time
8000
2.5
6000
4000
2000
0
7:30 PM
9:00 PM
10:30 PM
12:00 AM
Time
1:30 AM
3:00 AM
4:30 AM
6:00 AM
SCS Curve Number Method





Soil Conservation Service(SCS) Curve Number (CN) model
estimates precipitation excess as a function of cumulative
precipitation, soil cover, land use, and antecedent moisture
SCS developed the method for small basins (< 400 sq. mi.) to
"before" and "after" hydrologic response from events.
Classify soils (60 or 70 types) into four hydrologic soil groups
Method is simple enough to be used by people that have little
experience with hydrology.
Converts basin storage into something simpler and more
manageable (a “curve number” CN)
Abstractions – SCS Method

In general
Pe  P
After runoff begins
Fa  S

Potential runoff
P  Pe  I a  Fa
Precipitation

Pe
P  Ia

SCS Assumption
Fa
Pe

S
P  Ia

Solve for Rainfall Excess
P  I a 2
Pe 
P  Ia  S
Ia
Fa
tp
Time
P  Total Rainfall
Pe  Rainfall Excess
I a  Initial Abstraction
Fa  Continuing Abstraction
S  Potential Maximum Storage
SCS Method (Cont.)

Experiments showed


I a  0.2S
So
Pe 
12
P  0.2S 2
P  0.8S
1000
S
 10
CN
(American Units; 0  CN  100)
25400
 254CN
CN
(SI Units; 30  CN  100)
S

Impervious: CN = 100
Natural: CN < 100
100
90
80
70
11
Cumulative Direct Runoff, Pe, in

Surface
10
9
60
40
20
10
8
7
6
5
4
3
2
1
0
0
1
2
3
4
5
6
7
Cumulative Rainfall, P, in
8
9
10
11
12
SCS Method (Cont.)



S and CN depend on antecedent rainfall conditions
Normal conditions, AMC(II)
Dry conditions, AMC(I) CN ( I )  4.2CN ( II )
10  0.058CN ( II )

Wet conditions, AMC(III)
CN ( III ) 
23CN ( II )
10  0.13CN ( II )
5-day antecedent rainfall (in)
AMC Group
Dormant season
Growing season
I
< 0.50
< 1.4
II
0.5 -- 1.1
1.4 – 2.1
III
> 1.1
> 2.1
SCS Method (Cont.)

SCS Curve Numbers depend on soil conditions
Group
Minimum Infiltration
Rate (in/hr)
Soil type
A
0.3 – 0.45
High infiltration rates. Deep, well drained
sands and gravels
B
0.15 – 0.30
Moderate infiltration rates. Moderately
deep, moderately well drained soils with
moderately coarse textures
C
0.05 – 0.15
Slow infiltration rates. Soils with layers,
or soils with moderately fine textures
D
0.00 – 0.05
Very slow infiltration rates. Clayey soils,
high water table, or shallow impervious
layer
Example - SCS Method - 1



Rainfall: 5 in.
Area: 1000-ac
Soils:




Class B: 50%
Class C: 50%
Antecedent moisture: AMC(II)
Land use

Residential




Paved roads: 18% with curbs and storm sewers
Open land: 16%



40% with 30% impervious cover
12% with 65% impervious cover
50% fair grass cover
50% good grass cover
Parking lots, etc.: 14%
Example (SCS Method – 1, Cont.)
Hydrologic Soil Group
B
Land use
C
%
CN
Product
%
CN
Product
Residential (30% imp cover)
20
72
14.40
20
81
16.20
Residential (65% imp cover)
6
85
5.10
6
90
5.40
Roads
9
98
8.82
9
98
8.82
Open land: good cover
4
61
2.44
4
74
2.96
Open land: Fair cover
4
69
2.76
4
79
3.16
Parking lots, etc
7
98
6.86
7
98
6.86
40.38
50
Total
50
CN  40.38  43.40  83.8
43.40
Example (SCS Method – 1 Cont.)

Average AMC
CN  83.8
S
1000
 10
CN
1000
 10  1.93 in
83.8
Pe 

S
P  0.2S 2 5  0.2 *1.932
P  0.8S

5  0.8 *1.93
 3.25 in
Wet AMC
CN ( III ) 
S
23CN ( II )
23 * 83.8

 92.3
10  0.13CN ( II ) 10  0.13 * 83.8
1000
 10  0.83 in
92.3
Pe 
P  0.2S 2 5  0.2 * 0.832
P  0.8S

5  0.8 * 0.83
 4.13 in
Example (SCS Method – 2)


Given P, CN = 80, AMC(II)
Find: Cumulative abstractions and excess rainfall hyetograph
Time
(hr)
Cumulative
Rainfall (in)
P
0
0
1
0.2
2
0.9
3
1.27
4
2.31
5
4.65
6
5.29
7
5.36
Cumulative
Abstractions (in)
Ia
Fa
Cumulative
Excess Rainfall (in)
Pe
Excess Rainfall
Hyetograph (in)
Example (SCS Method – 2)






1000
1000
 10 
 10  2.50 in
CN
80
I a  0.2S  0.2 * 2.5  0.5 in
Calculate storage
Calculate initial abstraction
Initial abstraction removes
S
0.2 in. in 1st period (all the precip)
0.3 in. in the 2nd period (only part of
the precip)
Calculate continuing abstraction
Pe
Fa  S
P  Ia
P  Pe  I a  Fa
Time
(hr)
Cumulative
Rainfall (in)
P
0
0
1
0.2
2
0.9
S (P  I a )
2.5( P  0.5)
Fa 

(P  I a  S )
( P  2.0)
3
1.27
4
2.31
5
4.65
2.5(0.9  0.5)
Fa (2 hr) 
 0.34 in
(0.9  2.0)
6
5.29
7
5.36
Example (SCS method – 2)

Cumulative abstractions can now be calculated
Time
(hr)
Cumulative
Rainfall
(in)
Cumulative
Abstractions (in)
P
Ia
Fa
0
0
0
-
1
0.2
0.2
-
2
0.9
0.5
0.34
3
1.27
0.5
0.59
4
2.31
0.5
1.05
5
4.65
0.5
1.56
6
5.29
0.5
1.64
7
5.36
0.5
1.65
2.5( P  0.5)
Fa 
( P  2 .0 )
Example (SCS method – 2)
Cumulative excess rainfall can now be calculated
Excess Rainfall Hyetograph can be calculated


Time
(hr)
Cumulative
Rainfall (in)
P
0
0
1
0.2
2
0.9
3
1.27
4
2.31
5
4.65
6
5.29
7
5.36
R a i nf a l l ( i n)
R a i nf a l l H y e t ogr a phs
Cumulative
Abstractions (in)
2. 5
2
1. 5
1
0. 5
0
Pe  P  I a  Fa
Cumulative
Excess Rainfall (in)
Excess Rainfall
Hyetograph (in)
Ia
Fa
Pe
0
-
0
0
0.2
-
0
0
0.5
0.34
0.06
0.06
0.5
0.59
0.18
0.12
0.5
1.05
0.76
0.58
0.5
1.56
2.59
1.83
1.64
3.15
0.56
0.5
0.5
0
1
1.65
2
3
T i me ( h o u r )
4
3.21
5
0.06
6
7
R a i nf a l l
E x c e s s R a i nf a l l
Time of Concentration


Different areas of a watershed contribute to
runoff at different times after precipitation
begins
Time of concentration
Time at which all parts of the watershed begin
contributing to the runoff from the basin
 Time of flow from the farthest point in the
watershed
